Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{k - 1}{k^3 - 10k^2 + 9k} \times \dfrac{-5k^3 - 15k^2 + 350k}{k - 7} $
First factor out any common factors. $a = \dfrac{k - 1}{k(k^2 - 10k + 9)} \times \dfrac{-5k(k^2 + 3k - 70)}{k - 7} $ Then factor the quadratic expressions. $a = \dfrac {k - 1} {k(k - 1)(k - 9)} \times \dfrac {-5k(k - 7)(k + 10)} {k - 7} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {(k - 1) \times -5k(k - 7)(k + 10) } { k(k - 1)(k - 9) \times (k - 7)} $ $a = \dfrac {-5k(k - 7)(k + 10)(k - 1)} {k(k - 1)(k - 9)(k - 7)} $ Notice that $(k - 1)$ and $(k - 7)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-5k(k - 7)(k + 10)\cancel{(k - 1)}} {k\cancel{(k - 1)}(k - 9)(k - 7)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $a = \dfrac {-5k\cancel{(k - 7)}(k + 10)\cancel{(k - 1)}} {k\cancel{(k - 1)}(k - 9)\cancel{(k - 7)}} $ We are dividing by $k - 7$ , so $k - 7 \neq 0$ Therefore, $k \neq 7$ $a = \dfrac {-5k(k + 10)} {k(k - 9)} $ $ a = \dfrac{-5(k + 10)}{k - 9}; k \neq 1; k \neq 7 $